3.486 \(\int \frac {(c \sin (a+b x))^m}{\sqrt {d \sec (a+b x)}} \, dx\)

Optimal. Leaf size=77 \[ \frac {\sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac {1}{4},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1)} \]

[Out]

(cos(b*x+a)^2)^(1/4)*hypergeom([1/4, 1/2+1/2*m],[3/2+1/2*m],sin(b*x+a)^2)*(c*sin(b*x+a))^(1+m)*(d*sec(b*x+a))^
(1/2)/b/c/d/(1+m)

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Rubi [A]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2586, 2577} \[ \frac {\sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{m+1} \, _2F_1\left (\frac {1}{4},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(a+b x)\right )}{b c d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^m/Sqrt[d*Sec[a + b*x]],x]

[Out]

((Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, (1 + m)/2, (3 + m)/2, Sin[a + b*x]^2]*Sqrt[d*Sec[a + b*x]]*(c*S
in[a + b*x])^(1 + m))/(b*c*d*(1 + m))

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2586

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(1*(b*Cos[e +
f*x])^(n + 1)*(b*Sec[e + f*x])^(n + 1))/b^2, Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && LtQ[n, 1]

Rubi steps

\begin {align*} \int \frac {(c \sin (a+b x))^m}{\sqrt {d \sec (a+b x)}} \, dx &=\frac {\left (\sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)}\right ) \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^m \, dx}{d^2}\\ &=\frac {\sqrt [4]{\cos ^2(a+b x)} \, _2F_1\left (\frac {1}{4},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(a+b x)\right ) \sqrt {d \sec (a+b x)} (c \sin (a+b x))^{1+m}}{b c d (1+m)}\\ \end {align*}

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Mathematica [C]  time = 1.75, size = 289, normalized size = 3.75 \[ \frac {8 c (m+3) \sin ^2\left (\frac {1}{2} (a+b x)\right ) \cos ^4\left (\frac {1}{2} (a+b x)\right ) F_1\left (\frac {m+1}{2};-\frac {1}{2},m+\frac {3}{2};\frac {m+3}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) (c \sin (a+b x))^{m-1}}{b (m+1) \sqrt {d \sec (a+b x)} \left ((\cos (a+b x)-1) \left ((2 m+3) F_1\left (\frac {m+3}{2};-\frac {1}{2},m+\frac {5}{2};\frac {m+5}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+F_1\left (\frac {m+3}{2};\frac {1}{2},m+\frac {3}{2};\frac {m+5}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right )+(m+3) (\cos (a+b x)+1) F_1\left (\frac {m+1}{2};-\frac {1}{2},m+\frac {3}{2};\frac {m+3}{2};\tan ^2\left (\frac {1}{2} (a+b x)\right ),-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*Sin[a + b*x])^m/Sqrt[d*Sec[a + b*x]],x]

[Out]

(8*c*(3 + m)*AppellF1[(1 + m)/2, -1/2, 3/2 + m, (3 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b
*x)/2]^4*Sin[(a + b*x)/2]^2*(c*Sin[a + b*x])^(-1 + m))/(b*(1 + m)*(((3 + 2*m)*AppellF1[(3 + m)/2, -1/2, 5/2 +
m, (5 + m)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + AppellF1[(3 + m)/2, 1/2, 3/2 + m, (5 + m)/2, Tan[(a +
b*x)/2]^2, -Tan[(a + b*x)/2]^2])*(-1 + Cos[a + b*x]) + (3 + m)*AppellF1[(1 + m)/2, -1/2, 3/2 + m, (3 + m)/2,
Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + Cos[a + b*x]))*Sqrt[d*Sec[a + b*x]])

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{m}}{d \sec \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m/(d*sec(b*x + a)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin \left (b x + a\right )\right )^{m}}{\sqrt {d \sec \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^m/sqrt(d*sec(b*x + a)), x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin \left (b x +a \right )\right )^{m}}{\sqrt {d \sec \left (b x +a \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x)

[Out]

int((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin \left (b x + a\right )\right )^{m}}{\sqrt {d \sec \left (b x + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^m/sqrt(d*sec(b*x + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^m}{\sqrt {\frac {d}{\cos \left (a+b\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x))^m/(d/cos(a + b*x))^(1/2),x)

[Out]

int((c*sin(a + b*x))^m/(d/cos(a + b*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin {\left (a + b x \right )}\right )^{m}}{\sqrt {d \sec {\left (a + b x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))**m/(d*sec(b*x+a))**(1/2),x)

[Out]

Integral((c*sin(a + b*x))**m/sqrt(d*sec(a + b*x)), x)

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